show that $sum{n^frac{-3}{2}int_n^{2n}f(t)dt}=+infty $ [closed]
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$f :[0, +infty)to(0,+infty )$ continous and descending with $lim_{xto infty}f(x)=frac{1}{2}$ show that $$sum{n^frac{-3}{2}int_n^{2n}f(t)dt}=+infty $$
real-analysis sequences-and-series continuity
asked Nov 28 '18 at 13:30
stelioballstelioball
355
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closed as off-topic by Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin Nov 28 '18 at 19:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
$f :[0, +infty)to(0,+infty )$ continous and descending with $lim_{xto infty}f(x)=frac{1}{2}$ show that $$sum{n^frac{-3}{2}int_n^{2n}f(t)dt}=+infty $$
real-analysis sequences-and-series continuity
asked Nov 28 '18 at 13:30
stelioballstelioball
355
$endgroup$
closed as off-topic by Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin Nov 28 '18 at 19:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$f :[0, +infty)to(0,+infty )$ continous and descending with $lim_{xto infty}f(x)=frac{1}{2}$ show that $$sum{n^frac{-3}{2}int_n^{2n}f(t)dt}=+infty $$
real-analysis sequences-and-series continuity
asked Nov 28 '18 at 13:30
stelioballstelioball
355
$endgroup$
$f :[0, +infty)to(0,+infty )$ continous and descending with $lim_{xto infty}f(x)=frac{1}{2}$ show that $$sum{n^frac{-3}{2}int_n^{2n}f(t)dt}=+infty $$
real-analysis sequences-and-series continuity
real-analysis sequences-and-series continuity
asked Nov 28 '18 at 13:30
stelioballstelioball
355
asked Nov 28 '18 at 13:30
stelioballstelioball
355
edited Nov 28 '18 at 19:22
stelioball
asked Nov 28 '18 at 13:30
stelioballstelioball
355
asked Nov 28 '18 at 13:30
stelioballstelioball
355
asked Nov 28 '18 at 13:30
stelioballstelioball
355
355
closed as off-topic by Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin Nov 28 '18 at 19:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin Nov 28 '18 at 19:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, José Carlos Santos, Gibbs, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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$begingroup$
begin{align}
sum_{n=1}^Nn^{-frac 32}int_n^{2n}f(t)dt
&geqsum_{n=1}^Nn^{-frac 32}int_n^{2n}frac 12dt\
&=sum_{n=1}^Nn^{-frac 32}frac n2\
&=frac 12sum_{n=1}^Nfrac 1{sqrt n}\
&geqfrac 12sum_{n=1}^Nfrac 1n\
&xrightarrow{Ntoinfty}infty
end{align}
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
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$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
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I added a step in my answer.
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– Fabio Lucchini
Nov 28 '18 at 22:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
begin{align}
sum_{n=1}^Nn^{-frac 32}int_n^{2n}f(t)dt
&geqsum_{n=1}^Nn^{-frac 32}int_n^{2n}frac 12dt\
&=sum_{n=1}^Nn^{-frac 32}frac n2\
&=frac 12sum_{n=1}^Nfrac 1{sqrt n}\
&geqfrac 12sum_{n=1}^Nfrac 1n\
&xrightarrow{Ntoinfty}infty
end{align}
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
$endgroup$
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
add a comment |
$begingroup$
begin{align}
sum_{n=1}^Nn^{-frac 32}int_n^{2n}f(t)dt
&geqsum_{n=1}^Nn^{-frac 32}int_n^{2n}frac 12dt\
&=sum_{n=1}^Nn^{-frac 32}frac n2\
&=frac 12sum_{n=1}^Nfrac 1{sqrt n}\
&geqfrac 12sum_{n=1}^Nfrac 1n\
&xrightarrow{Ntoinfty}infty
end{align}
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
$endgroup$
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
add a comment |
$begingroup$
begin{align}
sum_{n=1}^Nn^{-frac 32}int_n^{2n}f(t)dt
&geqsum_{n=1}^Nn^{-frac 32}int_n^{2n}frac 12dt\
&=sum_{n=1}^Nn^{-frac 32}frac n2\
&=frac 12sum_{n=1}^Nfrac 1{sqrt n}\
&geqfrac 12sum_{n=1}^Nfrac 1n\
&xrightarrow{Ntoinfty}infty
end{align}
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
$endgroup$
begin{align}
sum_{n=1}^Nn^{-frac 32}int_n^{2n}f(t)dt
&geqsum_{n=1}^Nn^{-frac 32}int_n^{2n}frac 12dt\
&=sum_{n=1}^Nn^{-frac 32}frac n2\
&=frac 12sum_{n=1}^Nfrac 1{sqrt n}\
&geqfrac 12sum_{n=1}^Nfrac 1n\
&xrightarrow{Ntoinfty}infty
end{align}
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
edited Nov 28 '18 at 22:25
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
answered Nov 28 '18 at 13:38
Fabio LucchiniFabio Lucchini
7,81311426
7,81311426
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
add a comment |
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
how is it that $n^{frac{-3}{2}}>frac{1}{sqrt n}$
$endgroup$
– stelioball
Nov 28 '18 at 19:14
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
$begingroup$
I added a step in my answer.
$endgroup$
– Fabio Lucchini
Nov 28 '18 at 22:26
add a comment |
